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AB+AC+BC=50
AB=AC,BC=2BD
AB+BD=25
AD=40-25=15
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As shown in the figure, in △ ABC, ad is the center line on the BC side, and the difference between the circumference of △ abd and △ ADC is 5cm The sum of AB and AC is 19 cm (AB > AC). Find the length of AB and AC
∵ ad is the center line on the side of BC
∴BD＝CD
∵ △ abd perimeter = AB + BD + ad, △ ACD perimeter = AC + CD + ad
∴（AB+BD+AD）-（AC+CD+AD）＝5
∴AB-AC＝5
∴AB＝AC+5
∵AB+AC＝19
∴2AC+5＝19
∴AC＝7（cm）
∴AB＝7+5＝12（cm）
As shown in the figure, in the triangle ABC, ad is the center line on the BC side. The circumference of the triangle abd is 5 smaller than that of the triangle ACD. Can you find the difference between the edge lengths of AC and ab?
Yes
The circumference of △ abd = AB + BD + ad,
The circumference of △ ACD = AC + CD + ad,
And because ad is the center line on the side of BC,
So BD = CD
The circumference of ∵ △ abd is 5 times smaller than that of ∵ ACD,
∴AC+CD+AD-（AB+BD+AD）=AC-AB=5．
That is, the difference of side length between AC and ab is 5
In △ ABC, ad is the center line on the BC side, the perimeter of △ ADC is 3cm longer than that of △ abd, the sum of AB and AC is 11cm, and the length of AB side is?
According to the meaning of the title:
AD+DC+AC-AB-BD-AD=3
Because BD = DC, ac-ab = 3, AC + AB = 11
So AC = 7, ab = 4cm
As shown in the figure, we know: in △ ABC, BC ＜ AC, the vertical bisector de on the edge of AB intersects AB at D, intersects AC with E, AC = 9 cm, the circumference of △ BCE is 15 cm, calculate the length of BC
∵ de bisects AB vertically,
∴AE=EB，
The circumference of BCE is 15cm,
∴BC+EC+EB=15cm，
∵AC=EC+AE=9cm，
∴BC=15-9=6cm．
It is known that the circumference of △ ABC is 36cm, and ab = AC, and ad ⊥ BC, D is a vertical foot, and the circumference of ⊥ abd is 30cm, then ad =?
AD=12cm
2AB+BC=36
AB+1/2BC=18
AB+1/2BC+AD=30
AD=12
As shown in the figure, if the circumference of △ ABC is 32, and ab = AC, ad ⊥ BC is in D, and the circumference of ⊥ ACD is 24, then the length of ad is______ .
∵AB=AC，AD⊥BC，
∴BD=DC．
∵AB+AC+BC=32，
AB + BD + CD + AC = 32,
∴AC+DC=16
∴AC+DC+AD=24
∴AD=8．
Therefore, fill in 8
As shown in the figure, in △ ABC, ab = AC, BD ⊥ AC is proved to be ∠ a = 2 ∠ CBD at point D
Do AM ⊥ BC through A
AB=AC
∠BAM=∠MAC=1/2∠BAC
∠MAC+∠C=90°
∠C+∠DBC=90°、
∠MAC=∠DBC
∠BAC=2∠CBD
As shown in the figure, the diameter of ⊙ o is ab = 8cm, ﹤ CBD = 30 ° and the length of chord DC is calculated
Connect OC and OD, as shown in the figure,
∵∠DBC=1
2∠DOC，∠CBD=30°，
∴∠DOC=60°，
And OC=OD,
The △ cod is an equilateral triangle,
∴DC=OD，
And ∵ diameter AB = 8cm,
∴OD=4cm
So CD = 4cm
In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
∠ABC=180-∠CBD=180°90°=90°，
∠ACB=180°-∠A-∠ABC，
=180°-30°-90°，
=60°，
∠BCE=180°-∠ACB，
=180°-60°，
=120°．
A: BCE is 120 degrees
RELATED INFORMATIONS
- 1. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
- 2. It is known that AC and BD are two mutually perpendicular chords of the circle O: x2 + y2 = 4, and the perpendicular foot is m (1, 2) The maximum area of the quadrilateral ABCD is () A. 4 B. 4 Two C. 5 D. 5 Two
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- 8. 0
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- 17. As shown in the figure, both the triangle ABC and the triangle Dec are isosceles right triangles. The angle ACB = angle DCE = 90 ° f is the midpoint of De, h is the midpoint of AE, and G is the midpoint of BD If the △ Dec in Fig. 1 is rotated clockwise by an acute angle around point C to Fig. 2, is the triangle FGH still an isosceles right triangle
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